Which modification is usually tried first when a child is diagnosed with juvenile idiopathic arthritis (JI
Aspirin
Corticosteroids
Nonsteroidal anti-inflammatory drugs (NSAIDs)
Disease Modifying Anti-Rheumatoid Drugs (DMARDs)
The Correct Answer is C
The correct answer is c. Nonsteroidal anti-inflammatory drugs (NSAIDs)
Choice A reason:
Aspirin was once commonly used to treat juvenile idiopathic arthritis (JIA), but it is no longer the first-line treatment due to its potential side effects, such as gastrointestinal issues and Reye’s syndrome in children. While it can still be used in some cases, it is not the preferred initial treatment.
Choice B Reason:
Corticosteroids are effective in reducing inflammation and controlling symptoms of JIA, but they are not typically used as the first-line treatment due to their potential side effects, including weight gain, growth suppression, and increased risk of infections. They are usually reserved for more severe cases or when other treatments have failed.
Choice C Reason:
Nonsteroidal anti-inflammatory drugs (NSAIDs) are usually the first-line treatment for juvenile idiopathic arthritis. They help reduce inflammation, relieve pain, and improve joint function. NSAIDs are generally well-tolerated and have a long track record of safety and effectiveness in managing JIA.
Choice D Reason:
Disease Modifying Anti-Rheumatoid Drugs (DMARDs), such as methotrexate, are used in the treatment of JIA, but they are not typically the first-line treatment. DMARDs are often prescribed when NSAIDs are not sufficient to control the symptoms or when the disease is more severe. They help slow the progression of the disease and prevent joint damage.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is A
Explanation
Choice A reason:
Cardiac arrhythmia is a primary clinical manifestation of hyperkalemia. Hyperkalemia, defined as an elevated level of potassium in the blood, can significantly affect the electrical activity of the heart. This can lead to various types of arrhythmias, including bradycardia, ventricular tachycardia, and even cardiac arrest1. The presence of arrhythmias is a critical indicator of hyperkalemia and requires immediate medical attention to prevent life-threatening complications.
Choice B reason:
Seizures are not typically associated with hyperkalemia2. While severe electrolyte imbalances can potentially lead to neurological symptoms, seizures are more commonly linked to conditions such as hyponatremia (low sodium levels) or hypocalcemia (low calcium levels). Therefore, seizures are not a primary sign of hyperkalemia.
Choice C reason:
Dyspnea, or difficulty breathing, can occur in various medical conditions, including heart failure and respiratory disorders. While hyperkalemia can lead to muscle weakness and fatigue, which might indirectly affect breathing, dyspnea is not a primary clinical manifestation of hyperkalemia. The main concern with hyperkalemia is its effect on cardiac function.
Choice D reason:
Oliguria, or reduced urine output, is a symptom of acute renal failure but not specifically indicative of hyperkalemia. While acute renal failure can lead to hyperkalemia due to the kidneys’ inability to excrete potassium, oliguria itself is not a direct sign of hyperkalemia. The focus should be on the cardiac effects of elevated potassium levels
Correct Answer is B
Explanation
The correct answer is b. 25%.
Choice A: 50%
If both parents are heterozygous for the sickle cell trait (carriers), each child has a 50% chance of inheriting one sickle cell gene from one parent and a normal gene from the other parent. This would make the child a carrier of the sickle cell trait, not someone with sickle cell anemia. Therefore, the chance of having sickle cell anemia is not 50%.
Choice B: 25%
When both parents are carriers of the sickle cell trait (heterozygous), there is a 25% chance that their child will inherit two sickle cell genes (one from each parent), resulting in sickle cell anemia. This is because each parent has one normal hemoglobin gene (A) and one sickle cell gene (S). The possible combinations for their children are AA (normal), AS (carrier), SA (carrier), and SS (sickle cell anemia). The probability of the SS combination is 25%.
Choice C: 75%
A 75% chance is not accurate in this scenario. The 75% figure might be mistakenly considered if one were to add the probabilities of being a carrier (50%) and having sickle cell anemia (25%). However, these probabilities are distinct and should not be combined in this manner.
Choice D: 100%
A 100% chance would imply that every child of the couple would have sickle cell anemia, which is not the case. Since each parent is a carrier, there is only a 25% chance for each child to have sickle cell anemia. The remaining 75% of the time, the child will either be a carrier or have normal hemoglobin.
Whether you are a student looking to ace your exams or a practicing nurse seeking to enhance your expertise , our nursing education contents will empower you with the confidence and competence to make a difference in the lives of patients and become a respected leader in the healthcare field.
Visit Naxlex, invest in your future and unlock endless possibilities with our unparalleled nursing education contents today